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Transcript
Three blocks of masses M1=2 kg, M2=4 kg, and M3=6 kg are connected by strings on a frictionless inclined plane of 60 o, as shown in the figure below. A force of F=120N is applied upward along the incline, causing the system of three masses to accelerate up the incline. Consider the strings to be massless and taut. What is the acceleration of M2? Find the tensions T1 and T2 in the two sections of string. • • • • • • • • • Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 • • • • • • • Question 10 Question 11 Question 12 Question 13 Question 14 Question 15 Paired Problem 1 Notes * For simplicity when analyzing this system, we choose the the x-axis to be parallel to the surface of the incline and the y-axis to be perpendicular to the surface. Let’s define the positive direction as “upward” for both x and y. * Vectors quantities will be in bold type 1.Which one of these statements finishes the following sentence correctly? (there is no acceleration in the y-direction, so “a” refers to the acceleration in the x-direction throughout this tutorial) The acceleration of M2 is expected to be… • A : the same as the acceleration of the system as a whole. • B : less than the acceleration of the system as a whole. • C : greater than the acceleration of the system as a whole. Choice: A Correct Newton’s Second Law tells us: F=Ma where: M = M1+M2+M3 (total mass of the system) F is the net force on the system a is the acceleration of the whole system F = (M1+M2+M3 )a = M1a+M2a+M3a Thus, each block accelerates with the same “a,” which is also the acceleration of the system as a whole. Choice: B Incorrect Remember that the strings are all taut, which allows the masses to move together. Choice: C Incorrect Remember that the strings are all taut, which allows the masses to move together. 2. Which one of the following free body diagrams is correct for the system of three blocks? F A: Weight (W=Mg) Force of Friction (Ff) Normal Force (FN) FN F B: Mg Mg FN F C: FN F D: Ff Mg Mg (the length of the arrows does not represent the magnitudes of the forces here) Choice: A Incorrect It is necessary to include the normal force (FN). Choice: B Correct This diagram shows all of the forces acting on the system. Choice: C Incorrect The gravitational force always acts perpendicular to the ground, not perpendicular to the surface of the plane. Also, there is no friction between the blocks and the inclined plane. Choice: D Incorrect The gravitational force always acts perpendicular to the ground, not perpendicular to the plane of contact. 3. A good choice of coordinate axes may be to consider the x-axis parallel to the surface of the inclined plane and the y-axis perpendicular to it. (The positive x and y directions should be upwards.) Which forces are aligned with these axes? • A: FN and W • B: F and W • C: FN and F Choice: A Incorrect FN is in the positive y-direction (always perpendicular to the surface), but W is not along either coordinate axis. Remember that W=Mg, where g is gravitational acceleration, which always points downward (towards Earth). Choice: B Incorrect F is in the positive x-direction, but W is not along either coordinate axis. Remember that W=Mg, where g is the acceleration due to gravity, which always points downward (towards Earth). Choice: C Correct FN is directed in the positive ydirection and F is directed in the positive x-direction. 4. When we resolve the gravitational force into its x and y components, we get which one of the following? (the length of the arrows does not represent the magnitudes of the forces here) Mgcos() A: B: Mgsin() Mgsin() Mgcos) C: D: Mgsin() Mgcos() Mgcos() Mgsin() Choice: A Incorrect Mgcos() should point in the negative y-direction. Choice: B Incorrect Mgsin() should point in the negative x-direction. Choice: C Correct Here, the components are correctly resolved. Choice: D Incorrect Check the directions of both components. 5. Which one of the following is the correct equation of motion (x-component) for the whole system (using Newton’s Second Law)? Remember M=M1+M2+M3 • A: • B: • C: F=Ma F-Mgsin()=0 F-Mgsin()=Ma Choice: A Incorrect Don’t forget the effect of gravity Choice: B Incorrect The original problem statement tells us that the applied force is causing upward acceleration. Therefore, there is a net force; it is not equal to zero. Choice: C Correct This equation is correct according to Newton’s Second Law. The applied force on the system of blocks is in the opposite direction to the x-component of the weight of the blocks. We know from the original problem statement that the net force is not zero. 6. The acceleration of the system of blocks can be expressed by which of the following equations? F • A: a M F • B: a gsin(60 ) M • C: a F Mgsin (60 ) Choice: A Incorrect This equation is incorrect because it doesn’t take gravity into consideration. Choice: B Correct This is what we get when we rearrange the equation that we found for force in question 6. F Mgsin( 60) Ma F gsin( 60) a M M M M Choice: C Incorrect This is just the net force on the system. We are looking for the acceleration of the system. 7. Determine the value of the acceleration a of the blocks up the incline. Reasoning: F a gsin( 60 ) M Where: F=120N M= M1+M 2+M3= 2 kg +4 kg+6 kg= 12 kg g= 9.8 m/s2 120N a (9.8m/s2 )(0.866) 1.5 m/s2 12kg • As discussed earlier, the acceleration of M2 is the same as that of the system as a whole because the blocks move together. • Now let’s focus on finding the tensions T1 and T2. 8. Which one of these is the free body diagram of M2? (the length of the arrows does not represent the magnitudes of the forces here) A: FN T1 C: F+T2 T2 M2g T2 B: T1 M2 g FN M1 g FN D: FN T2 M2 g T1 M2g Choice: A Incorrect F only acts on M3. Choice: B Correct This diagram shows all of the forces acting upon M2. Choice: C Incorrect The gravitational force on M1 does not act on M2. Therefore, it does not belong in the free body diagram for M2. Include T1. Choice: D Incorrect The force on an object from contact with a string in tension always acts in the direction away from the object. The directions of T1 and T2 should be switched. 9. Which of the following is the correct representation of the x-component of Newton’s second law? You must resolve the gravitational force into components. • A : T T1 M2a • B : T T1 M2gsin( 60 ) M2a • C : T M2a 2 2 2 Choice: A Incorrect This expression does not include the gravitational force. Choice: B Correct This expression is consistent with the free body diagram. Choice: C Incorrect Gravitational force and tension T1 should also be considered because they are acting on M2. 10. We have two unknowns, T1 and T2. We need two equations. We have one equation from the free body diagram of M2. Get the other one from the free body diagram of M1. The free body diagram for M1 is: A: FN T1 B: T1+T2 FN M1 g M1 g C: FN F D: M1 g FN T1 M1g (the length of the arrows does not represent the magnitudes of the forces here) Choice: A Correct All of the forces are shown correctly. Choice: B Incorrect T2 doesn’t act on M1. Choice: C Incorrect F only acts on M3. Choice: D Incorrect The normal force acts perpendicular to the surface of the plane of contact. 11. Resolving the gravitational force into its components and applying Newton’s Second Law gives us which of the following as the xcomponent of the equation of motion for M1? • A : T1 M1a • B : T1 M1gsin( 60 ) M1a • C : T1 M1g M1a Choice: A Incorrect T1 is not the net force. Remember the force in Newton’s Second Law is the net force. Gravity acts upon M1. Choice: B Correct This equation is consistent with the free body diagram for M1. Choice: C Incorrect This would be correct if the angle of inclination were 90 degrees. 12. Which equation should be solved first? • A : T2 T1 M2gsin( 60 ) M2a • B : T1 M1gsin( 60 ) M1a Choice: A Incorrect We start with the other equation because this one contains two unknowns. It is not possible to solve for two unknowns using a single equation. Choice: B Correct This equation should be solved first since only T1 is unknown. 13. After solving for T1 you will get : T1 M1gsin( 60 ) M1a If you plug this back into the equation of motion for M2( T2 T1 M2gsin( 60 ) M2a ) and simplify, you get which of the following expressions for T2? • A : T2 M1gsin( 60 ) M2a • B : T2 M1a M2a • C : T2 (M2 M1)(gsin( 60 ) a) Choice: A Incorrect This expression omits several terms. Check your algebra. Choice: B Incorrect This expression does not include the contribution of gravity. Choice: C Correct We get this result using simple algebra and factoring out common variables. Now all you have to do is insert the given values and calculate the numerical values of T1 and T2. 14) Find the value of T1 Answer The correct answer is: T1 M1(a gsin( 60 )) 2kg(1.51m/s (9.8m/s )(0.866)) 2 20.0 N 2 15) Now find T2. Answer The correct answer is : T2 T1 M2gsin( 60 ) M2a T2 20N 4kg9.8m/s2 sin 60 4kg1.5m/s2 T2 60.0 N Paired Problem 1) Three blocks of masses M1=2 kg, M2=4 kg, and M3=6 kg are connected by strings on a frictionless inclined plane of 60 o, as shown in the figure below. A force of F=120N is applied upward along the incline, causing the system of three masses to accelerate up the incline. Consider the strings to be massless and taut. What is the acceleration of M2? Find the tensions T1 and T2 in the two sections of string.